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Entries in a row are separated by commas  (or plain spaces). Rows are separated by semi colon DEFINE RIGHT HAND SIDE VECTOR » I=[iA;-iA;iB]; %end in ";" to skip echo SOLVE LINEAR EQUATION » V=G\I % end with carriage return and get the echo V = 11.9940 15.9910 15.9940
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EARNING EXTENSION: FIND NODE VOLTAGES 0 10 4 10 : @ 2 1 1 1 = - + - k V V mA k V V NODE EQUATIONS 0 10 2 10 : @ 2 1 2 2 = + + - k V I k V V V O CONTROLLING VARIABLE (IN TERMS ON NODE VOLTAGES) k V I O 10 1 = REPLACE 0 10 10 2 10 0 10 4 10 2 1 1 2 2 1 1 = + + - = - + - k V k V k V V k V V mA k V REARRANGE AND MULTIPLY BY 10k 0 2 ] [ 40 2 2 1 2 1 = + = - V V V V V eqs.   add   and    2 / * V V V V 16 80 5 1 1 = = V V V V 8 2 2 1 2 - = - =
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O   V  VOLTAGE THE   FIND NODE EQUATIONS 0 6 3 2 = + + - k V k V mA x x NOTICE REPLACEMENT OF DEPENDENT SOURCE  IN TERMS OF NODE VOLTAGE 0 12 12 6 = + + - k V k V k V O O x k 6 / * k 12 / * ] [ 4 0 2 2 ] [ 4 ] [ 12 3 V V V V V V V V O x O x x = = - = = LEARNING EXTENSION
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3 nodes plus the reference. In  principle one needs 3 equations.. …but two nodes are connected to the reference through voltage  sources. Hence those node  voltages are known!!! …Only one KCL is necessary 0 12 12 6 1 2 3 2 2 = - + - + k V V k V V k V ] [ 5 . 1 ] [ 6 4 0 ) ( ) ( 2 2 2 1 2 3 2 2 V V V V V V V V V = = = - + - + EQUATIONS   THE   SOLVING Hint: Each voltage source connected to the reference node saves one node equation One more example …. IRCUITS WITH INDEPENDENT VOLTAGE SOURCES ] [ 6 ] [ 12 3 1 V V V V - = = THESE ARE THE REMAINING TWO NODE EQUATIONS
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+ - 2 S I 3 S I 1 S V 1 S I 1 R 2 R 3 R 4 R Problem 3.67 (6th Ed) Find V_0 R1 = 1k; R2 = 2k, R3 = 1k, R4 =  2k Is1 =2mA, Is2 = 4mA, Is3 = 4mA, Vs = 12 V IDENTIFY AND LABEL ALL NODES WRITE THE NODE EQUATIONS ] [ 12 : @ 3 3 V V V V VS = =  VOLTAGE NODE    KNOWN 0 2 1 ] [ 2 0 : @ 1 2 1 4 1 1 2 1 1 1 = + - + - = + - + - k V k V V mA R V R V V I V S 0 2 1 12 1 ] [ 4 0 : @ 4 2 2 1 2 2 4 2 3 3 2 1 1 2 3 2 = - + - + - + - = - + - + - + - k V V k V k V V mA R V V R V V R V V I V S 0 2 ] [ 4 ] [ 2 0 : @ 2 4 2 2 4 2 1 4 = - + - = - + - k V V mA mA R V V I I V S S NOW WE LOOK WHAT IS BEING  ASKED TO DECIDE THE SOLUTION STRATEGY. 2 1 0 V V V - = 1 V 2 V 3 V 4 V - + O V     FOR   NEEDED   ARE ONLY  O V V V 2 1 , - +
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0 2 1 ] [ 2 1 2 1 = + - + - k V k V V mA 0 2 1 12 1 ] [ 4 4 2 2 1 2 = - + - + - + - k V V k V k V V mA 0 2 ] [ 4 ] [ 2 2 4 = - + - k V V mA mA TO SOLVE BY HAND ELIMINATE DENOMINATORS ] [ 4 2 3 2 1 V V V = - */2k */2k ] [ 32 5 2 4 2 1 V V V V = - + - */2k ] [ 4 4 2 V V V = + - Add 2+3 ] [ 36 4 2 2 1 V V V = + - ] [ 10 ] [ 40 4 1 1 V V V V = = ] [ 14 ] [ 56 4 2 2 V V V V = = FINALLY!! ] [ 4 2 1 0 V V V V - = - = = - - - - 4 32 4 1 1 0 1 5 2 0 2 3 3 2 1 V V V ALTERNATIVE: USE LINEAR ALGEBRA (1) (2) (3) So. What happens when sources are connected between two  non reference nodes? ] [ 4 2 3 2 1 V V V = - add   and   2 / *
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 will use this example to introduce the concept of a SUPERNODE Conventional node analysis  requires all currents at a node @V_1 0 6 6 1 = + + - S I k V mA @V_2 0 12 4 2 = + + - k V mA I S  eqs, 3 unknowns...Panic!!  he current through the source is not elated to the voltage of the source ath solution: add one equation ] [ 6 2 1 V V V = - Efficient solution Efficient solution : enclose the  source, and all elements in  parallel, inside a surface.
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  • Spring '12
  • HaroldKlee
  • Amplifier, Volt, Trigraph, SEPTA Regional Rail, Jaguar Racing

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